Ben Gurion University of the Negev
www.bgu.ac.il/atomchip
Physics 2B for Materials and Structural Engineering
Lecturer: Daniel Rohrlich
Teaching Assistants: Oren Rosenblatt, Shay Inbar
Week 7. Sources of the magnetic field – Biot-Savart law • parallel
currents • Ampère’s law • Gauss’s law for B • towards Maxwell’s
equations
Source: Halliday, Resnick and Krane, 5th Edition, Chap. 33.
Biot-Savart law
The last lecture was all about how moving charges respond to
magnetic fields. This lecture is all about how moving charges
create magnetic fields!
In 1820, the Danish scientist Hans Christian Oersted gave a
lecture-demonstration about how an electric current in a wire
generates heat. For another part of the demonstration, he had a
compass nearby. While lecturing, he noticed to his surprise
that every time the electric current was switched
on, the compass needle moved. He said nothing
but, after a few months of experiments, published
his discovery (in Latin) with no explanation.
Biot-Savart law
Within months, Jean-Baptiste Biot and Félix Savart had made
careful quantitative experiments on the force exerted by an
electric current on a nearby magnet. They summarized their
results in the law named after them.
dB
r
ds
Biot-Savart law
Each current element ds contributes to the magnetic field B(r)
0 I ds  rˆ
a contribution dB 
, where μ0 = 4π × 10–7 T·m/A .
4 r 2
dB
r
“the permeability (‫ חלחלות‬,‫ )פרמאביליות‬of free space”
ds
Biot-Savart law
Each current element ds contributes to the magnetic field B(r)
0 I ds  rˆ
a contribution dB 
, where μ0 = 4π × 10–7 T·m/A .
4 r 2
dB
r
• The vector dB is perpendicular to ds and to r.
• The vector dB is proportional to I, to 1/r2 and to |ds|.
• When ds and r are parallel, dB vanishes.
• When ds and r are not parallel, dB is perpendicular
to both of them, according to the right-hand rule,
and proportional to the sine of the ds → r angle.
ds
Right-hand rule (for the Biot-Savart law):
ds
r
dB
Biot-Savart law
Example 1: A point makes angles θ1 and θ2 with the ends of a
straight wire carrying current I. The distance of the point from
the wire is a. What is B?
y
Answer: Introduce x- and y-axes. Note that
r = (0,a) – (x,0) = (–x, a), that a = r sin θ and
thus r = a/sin θ. Note also that x = –r cos θ =
–a cot θ, so dx = a dθ /sin2 θ.
dB
a
θ1
r
θ
ds = dx
θ2
x
Biot-Savart law
Example 1: A point makes angles θ1 and θ2 with the ends of a
straight wire carrying current I. The distance of the point from
the wire is a. What is B?
y
Answer: The magnetic field is the integral
dB
a
θ1
0
B(r)  dB  I
4

0
ˆ
zI
2
4
r
but dx/r2 = dθ/a, so

ds  rˆ

sin 
r
2
dx
r
θ
ds = dx
θ2
x
Biot-Savart law
Example 1: A point makes angles θ1 and θ2 with the ends of a
straight wire carrying current I. The distance of the point from
the wire is a. What is B?
y
Answer: The magnetic field is the integral
0 2
B(r )  zˆ I
sin  d
4 a 1
dB

 zˆ I
a
θ1
r
θ
ds = dx
0
cos1  cos 2  .
4 a
θ2
x
Biot-Savart law
Example 2: What is B if the wire is straight and infinitely
long?
Answer: Then θ1 = 0 and θ2 = π so
0
cos1  cos 2 
B(r )  zˆ I
4 a
y
dB
 zˆ I
a
θ1
0
.
2 a
r
θ
ds = dx
θ2
x
Biot-Savart law
In this example, a different version of the right-hand rule is
useful: If your thumb shows the direction of the current, your
fingers curl in the direction of the magnetic field.
y
dB
a
θ1
r
θ
ds = dx
θ2
x
Biot-Savart law
Example 3: What is B a distance x along the symmetry axis of
a circular loop of radius R carrying current I?
Answer: The magnitude of dB is dB = μ0 I ds/4πr2 , where
r2 = x2 + R2, but only the x-component of dB survives. We
multiply dB by cos θ = R/r and replace ds by 2πR to get
B
y
r
R
θ
 0 IR 2
2 [x  R ]
2
2 3/ 2
.
dB
θ
x
Parallel currents
Now let’s look at magnetic fields and moving charges as a
complete system. Our first question is simple but important:
Two straight, infinitely long parallel wires, of separation a,
carry currents I1 and I2 . What is their mutual force per unit
length?
The field due to one current (say, I1) is μ0 I1/2πa. It exerts a
magnetic force (μ0 I1/2πa) I2L. on any section of length L in the
other wire. Therefore the magnetic force per unit length FB /L
is FB /L= μ0 I1I2/2πa. This force per unit length is invariant if
we interchange I1 and I2. If we check the direction of each
force, we find that the mutual forces are equal and opposite, as
Newton’s third law requires.
Parallel currents
Do parallel currents attract or repel each other? What about
anti-parallel currents?
L
I1
B2
F2
a
a
I2
Parallel currents
A large current begins to flow through a spring. Does the
spring expand or contract?
Parallel currents
We calculated the magnetic force per unit length to be
FB  0 I1I 2

L
2a
.
Now let’s try some numbers. Assume I1 = I2 = I, a = 1 m, and
FB/L = 2 × 10–7 N/m. Since μ0 = 4π × 10–7 T·m/A, we have
2 10 7 N / m  2 10 7 (T/A) I 2 .
Since 1 T = N /A·m (recall FB= qv × B), we obtain I2 = 1 A2
and an operational definition of the ampere:
When the force per unit length between two long, parallel
wires separated by 1 m and carrying identical currents I is
FB = 2 × 10–7 N/m, the current in each wire is 1 A.
Parallel currents
We likewise obtain an operational definition of the coulomb:
When a wire carries a steady current of 1 A, the amount of
charge that flows through the wire in 1 s is 1 C.
Ampère’s law
Oersted’s discovery that an electric current deflects a compass
needle suggests the following experiment:
I
B
dr
I=0
B(r) depends only on r. From the Biot-Savart law, B = μ0 I/2πr

and so in this case, at least, B(r )  dr  0 I .
Ampère’s law
Oersted’s discovery that an electric current deflects a compass
needle suggests the following experiment:
I
B
dr
I=0
Ampère discovered that this law holds in general, and the

equation B(r )  dr  0 I is called Ampère’s law.
Ampère’s law
Example 1: Rank  B(r )  dr for the four closed curves a, b, c, d
using Ampère’s law.
d
1A
5A
c
a
b
2A
Ampère’s law
Example 2: What is B inside and outside an infinitely long
conducting wire of radius R, carrying uniform current I?
Answer: By symmetry, B must circulate around the axis of the
wire. Outside the wire, we choose a concentric circle with
radius r > R. Then Ampère’s law tells us that 2πrB = μ0 I so
B = μ0 I/2πr.
B
I
Ampère’s law
Example 2: What is B inside and outside an infinitely long
conducting wire of radius R, carrying uniform current I?
Answer: By symmetry, B must circulate around the axis of the
wire. Inside the wire, we choose a concentric circle with radius
r < R. Then Ampère’s law tells us that 2πrB = μ0 I(πr2/πR2) so
B = μ0 Ir/2πR2.
B
I
Ampère’s law
Example 3: What is B for an infinite current sheet carrying
surface current density Js?
Answer: We use the rectangular
loop shown to evaluate
B
Js
B
 B(r)  dr .
The top and bottom sides don’t
contribute, while the right and left
sides each contribute BL, where L is
the height of the rectangle. We
have 2BL =μ0 JsL, so B =μ0 Js/2.
Ampère’s law
Example 4: Ampère’s law, like Gauss’s law, is especially
useful in cases of high symmetry. Let’s apply Ampère’s law
to find the field of a long solenoid, which would be awkward
using the Biot-Savart law.
Experiment shows that B
outside such a solenoid is
very weak.
W
2
What is the field B inside
the solenoid?
1
3
4
B
L
Ampère’s law
Example 4: Ampère’s law, like Gauss’s law, is especially
useful in cases of high symmetry. Let’s apply Ampère’s law
to find the field of a long solenoid, which would be awkward
using the Biot-Savart law.
Consider the integral of B·dr
around a rectangular path as
shown, in the direction shown.
On Side 3 , B is negligible. On
Sides 2 and 4 the integrals of
B·ds sum to 0. So B·dr on Side
1 is what must equal μ0 times
the current through the rectangle.
W
2
1
3
4
B
L
Ampère’s law
Example 4: Ampère’s law, like Gauss’s law, is especially
useful in cases of high symmetry. Let’s apply Ampère’s law
to find the field of a long solenoid, which would be awkward
using the Biot-Savart law.
We have BL = μ0 NI, where N
windings, each carrying current
I, pass through the rectangle.
Letting n = N/L denote the number
of windings per unit length, we
conclude that B = μ0 nI. This is
the magnetic field anywhere
inside the solenoid; it is uniform.
W
2
1
3
4
B
L
Gauss’s law for B
We have seen that the field lines of E and B are similar. One
striking difference is that there are no magnetic monopoles.
This similarity of the field lines is not just a coincidence. We
will see that the fundamental equations for E and B are very
similar.
First, there is a version of Gauss’s law for the magnetic field. It
states that total magnetic flux out of a closed surface equals 0:

 B  B  dA  0 .
This law can be derived from the law of Biot and Savart.
Towards Maxwell’s equations
The fundamental equations for E and B are very similar.
First, there is a version of Gauss’s law for the magnetic field. It
states that total magnetic flux out of a closed surface equals 0:

 B  B  dA  0 .
Second, we have already seen a statement about E that is very
similar to Ampère’s law:
 E(r)  dr  0
.
Remember Faraday?
Now we can add a new basic rule about
field lines:
•They never start or stop in empty space –
they stop or start on a charge or extend to
infinity.
•They never cross – if they did, a small
charge placed at the crossing would show
the true direction of the field there.
• The density of field lines in one
direction is proportional to the strength of
the field in the perpendicular direction.
• There are no closed loops!
E
E
E
E
Towards Maxwell’s equations
The set of four fundamental equations for E and B,

E  dA 
q
0
 E  dr  0
 B  dr  0 I
 B  dA  0 ,
(Gauss’s law)
(Ampère’s law)
together with the Lorentz force law FEM = q (E + v × B), sum
up everything we have learned so far about electromagnetism!
Towards Maxwell’s equations
The set of four fundamental equations for E and B,

E  dA 
q
0
 E  dr  0
 B  dr  0 I
 B  dA  0 ,
(Gauss’s law)
(Ampère’s law)
are correct for all static charges and steady-state currents, i.e.
for electrostatics and magnetostatics. But if E and B change in
time, two of the equations need correction. Guess which two?
Halliday, Resnick and Krane, 5th Edition, Chap. 33, Prob. 9:
A long solenoid, centered along the z-axis, has 100 turns per
centimeter. An electron moves within the solenoid in a circle
of radius 2.30 cm in the xy-plane. The speed of the electron is
0.0460c, where c = 3.00 × 108 m/s is the speed of light. What
is the current I in the wire of the solenoid?
Halliday, Resnick and Krane, 5th Edition, Chap. 33, Prob. 9:
A long solenoid, centered along the z-axis, has 100 turns per
centimeter. An electron moves within the solenoid in a circle
of radius 2.30 cm in the xy-plane. The speed of the electron is
0.0460c, where c = 3.00 × 108 m/s is the speed of light. What
is the current I in the wire of the solenoid?
Answer: The cyclotron frequency ω of the electron is
ω = v/r = 0.0460c/0.0230 m = 6.00 × 108 Hz
and we derived ω = eB/m where e is the electron charge and m
is its mass. Also B = μ0 nI for the field in a solenoid, hence I =
mω/neμ0 . Putting in the numbers (m = 9.11× 10–31 kg and e =
1.60 × 10–19 C) we obtain I = 0.272 A.
The problem with Ampère’s law
How do we apply Ampère’s law,
S
electrical circuit?
Current flows
through Surface S1.
No current flows
through Surface S2.
 B  dr  0  J  dA, to this
S
–q
∂S
B
q
r
Hence we have a
μ0I/2πr = B = 0:
CONTRADICTION!
Surface S2
I
Surface S1
The problem with Ampère’s law
You may say, “This is silly. Ampère’s law is not like Faraday’s
law, no matter what Ampère thought. Forget about S1 and S2!
Just consider how much current flows through the closed loop
∂S .”
Yes, but suppose the closed loop straddles the capacitor:
q
–q
I
∂S
How much current is flowing through the loop ∂S now?
Maxwell’s fix
This set of four fundamental equations for E and B,
 E  dA   0
q
(Gauss’s law)
 B  dA  0

d
E  dr    B
dt

d
B  dr   0 I   0  0  E ,
dt
is called Maxwell’s equations!
(Faraday’s law)
(Ampère’s law
as modified by
Maxwell)
Maxwell’s fix
Let’s check that Maxwell’s modification of Ampère’s law,

d
B  dr   0 I   0 0  E ,
dt
fixes the problem, for
the case of a parallelplate capacitor.
–q
∂S
B
q
r
If we consider
Surface S1, we have
(2πr)B = μ0I .
Surface S2
I
Surface S1
Maxwell’s fix
Let’s check that Maxwell’s modification of Ampère’s law,

d
B  dr   0 I   0 0  E ,
dt
fixes the problem, for
the case of a parallelplate capacitor.
If we consider
Surface S2, we have
(2πr)B = μ0ε0 dΦE/dt.
For a parallel-plate
capacitor, it still
equals μ0I!
–q
∂S
B
q
r
Surface S2
I
Surface S1
Maxwell’s fix
Let’s check that Maxwell’s modification of Ampère’s law,

d
B  dr   0 I   0 0  E ,
dt
fixes the problem, for the case of a parallel-plate capacitor.
If we consider Surface S1, we have (2πr)B = μ0I .
If we consider Surface S2, we have (2πr)B = μ0ε0 dΦE/dt.
For a parallel-plate capacitor of area A and plate separation d,
ΦE = AE = AV/d = Aq/Cd, and C = ε0 A/d, so ΦE = qA/Cd = q/ε0.
We get dΦE/dt = I/ε0 so (2πr)B = μ0ε0 dΦE/dt = μ0I!
Note I = ε0 dΦE/dt is often called the displacement current.